What a fluid with no length in its equations can do; and a prediction made in 1941, tested against twenty-five photographs of 1945.
Write out the equations of Module 2 and look for a length in them. There is none. Continuity, Euler and the energy equation contain densities, velocities, pressures and derivatives, and not one intrinsic scale — the viscosity and conductivity that would supply one were dropped in Module 2 on the strength of Module 1's Knudsen numbers. Every module since has treated that as a licence to work with smooth fields. This one treats it as a licence to do the opposite.
If the equations contain no length, nothing in them forbids a solution in which $\rho$, $v$ and $p$ change by a finite amount across zero distance. Such a surface is a shock. The physical layer is of course not of zero thickness: Problem D2 works out that in sea-level air it is a few mean free paths, about $2\times10^{-5}\ \mathrm{cm}$, which is $2\times10^{8}$ times thinner than the Trinity fireball was one millisecond after detonation. Against every other length in the problem that is zero, and the idealisation is safe by eight orders of magnitude.
The gain from the idealisation is larger than it first appears. Because the layer is unresolved, no equation of motion inside it is available — and none is needed. The three conservation laws, integrated across the layer, relate the state on one side to the state on the other with nothing said about the constitutive physics of the interior at all. That is §2, and it is the whole of shock theory in three lines. What comes out is not a small correction: the downstream gas can be a hundred times hotter than the upstream gas, and the ratio is fixed by the upstream Mach number alone.
A second absence of length drives the rest of the module. Deposit an energy $E$ at a point, instantaneously, in a uniform gas of density $\rho_0$, and ask for the radius $R(t)$ of the shock that runs out from it. The problem contains $E$, $\rho_0$ and $t$ and nothing else — no length, again. There is exactly one length that can be built from those three, so $R$ must be proportional to $(Et^2/\rho_0)^{1/5}$ and the exponent $2/5$ is settled before any dynamics is done. That argument is §4, and G. I. Taylor made it in 1941.
Four years later the argument met a test that no theoretician arranges for himself. On 16 July 1945 the first nuclear device was fired in New Mexico, and the fireball was photographed by a high-speed camera with a length scale in the frame. In 1950 Taylor published twenty-five radii against twenty-five times, read off that film, and compared them with the prediction he had made nine years before seeing it. §5 repeats the comparison from his own table, and finds two things: the exponent is confirmed to better than two and a half per cent over a factor of 620 in time, and the model that produced the exponent is refuted by the residuals from the very same fit. Both statements come from one table, and the second is the more instructive.
The module then leaves the laboratory twice. §6 uses a shock that a spacecraft flew through and measured from inside — Voyager 2 at the solar-wind termination shock in 2007, with Neptune's bow shock from the same instrument eighteen years earlier as a control. The control confirms the compression ceiling exactly; the termination shock refutes the single-fluid closure this book has used since Module 2, and the way it fails is precisely the way Module 1 said a collisionless plasma would fail. §7 then follows a supernova remnant from the moment the Sedov solution becomes valid to the moment it stops being valid, and §8 pays a debt: Module 7 closed by recording that its Rayleigh–Taylor half had no astronomical check anywhere in Modules 1–10. It has one now, and the check refutes the transfer.
Everything in this module is a statement about the configuration defined next, and every hypothesis it carries is listed once here so that the propositions can name it rather than restate it.
Three hypotheses there will be paid for later, and it is worth naming which section pays. Steady in the surface frame fails when the shock accelerates on a timescale comparable to the crossing time; it does not fail for any case in this module. Polytropic with the same $\gamma$ on both sides is the assumption §5 finds broken at Trinity, where dissociation, vibration and radiation all act behind the front and Taylor argues — with evidence, and not quite successfully — that they cancel. One fluid, one temperature is the assumption §6 finds broken at the termination shock, where the energy goes to a population of pickup ions that the single-fluid description has no room for.
Fig. 1 is the whole proof in a picture, and the question mark in the middle of it is the content of the proposition.
Three remarks on what the proof used, and one on what it did not. It required each equation to be a divergence, which is why the conservation forms were taken and not the convective ones: $\partial_t(\rho v)$ written as $\rho[\partial_t v + (v\cdot\nabla)v]$ is not a divergence and cannot be integrated across a discontinuity. It required the endpoints to lie in gas with no gradients, which is the whole working content of "the thickness is negligible": the layer must be thin enough that uniform gas is reached on both sides before anything else in the problem has changed. And it required every source to be a bounded volume source, so a layer that radiates away a finite amount of energy per unit area — a surface loss, not a bounded volume loss — breaks (2.4). That is the isothermal shock of Proposition 4, where (2.4) is deliberately replaced. What the proof did not use is any value of $\sigma_{nn}$, of $q_n$ or of the thickness, any equation of state inside the layer, or any assumption that the gas in there is in local thermodynamic equilibrium. That, and only that, is the sense in which the interior is not modelled.
The three fluxes are stated here for a gas with no magnetic field. A magnetised shock carries two further jump conditions, on the normal and tangential components of $\mathbf B$, and the momentum flux acquires the magnetic terms of Module 2. §6 uses a shock that is magnetised and quasi-perpendicular, and it is careful to compare only the quantities for which the difference does not matter; the place where it would matter is named there and the comparison is refused.It is worth pausing on how little was assumed. Nothing was said about viscosity, about heat conduction, about how thick the layer is or about whether the gas inside it is even in local thermodynamic equilibrium. All of that is real, and all of it is invisible to (2.2)–(2.4). What the conservation laws give up in exchange is any information about the structure of the transition: they say where the gas ends up, never how it got there. Problem D2 shows that the trade is a good one in air and a very bad one at the termination shock, where the layer is $10^5$ to $10^6$ times thinner than a collisional mean free path and the phrase "local thermodynamic equilibrium" has no referent at all.
Proposition 1 is three equations in the three downstream unknowns $\rho_2$, $v_2$, $p_2$, given the upstream state. Solving them is algebra, and the answer is worth more than the algebra.
The two behaviours in Proposition 2 are physically different and the contrast is the point of Fig. 2. Compression is bounded because the gas that piles up behind the front pushes back, and past a certain density no increase in the ram pressure can make it denser; heating is unbounded because the kinetic energy that has to go somewhere goes into internal energy, and internal energy has no ceiling. A strong shock is a device for making a hot gas, not a dense one. That single sentence explains why supernova remnants shine in X-rays rather than looking like dense shells, and why §7 can speak of $10^8\ \mathrm{K}$ gas at a density of a few particles per cubic centimetre.
The ceiling is approached quickly. At $\gamma = 5/3$, a Mach 3 shock has reached 75.00 per cent of it, Mach 5 gives 89.29 per cent, Mach 10 gives 97.09 per cent and Mach 30 gives 99.67 per cent. Problem D1 asks for the Mach number at which 99 per cent is reached and gets $M_1 = 17.2$ for $\gamma = 5/3$ and $22.2$ for $\gamma = 7/5$; the Trinity shock at one millisecond was at $M_1 = 43$ and was therefore strong by any measure.
Nothing in Proposition 1 prefers one direction. Put $M_1 \lt 1$ into (3.1) and the formulae return perfectly finite ratios: at $M_1 = 0.7$ and $\gamma = 5/3$ they give $\rho_2/\rho_1 = 0.5616$ and $p_2/p_1 = 0.3625$, a jump from dense to rarefied that satisfies mass, momentum and energy conservation exactly. Such a thing is never seen. The reason is not in the conservation laws, and this is the only place in Modules 1–8 where the selection is made by thermodynamics rather than by dynamics.
A second obstruction sits below the first, and it is worth separating because it is not thermodynamic at all. The pressure ratio in (3.1) vanishes at $M_1 = \sqrt{(\gamma-1)/2\gamma}$, which is $0.4472$ for $\gamma = 5/3$, and goes negative below it — at $M_1 = 0.4$ it returns $p_2/p_1 = -0.0500$. Below that Mach number the "rarefaction shock" is not a solution that the second law forbids; it is not a solution at all, because it would require a negative pressure. So the two objections are of different kinds and cover different ranges, and Problem C3 asks the reader to state both.
What happens instead is a rarefaction fan: an expansion spreads out into a continuous region whose width grows with time, because each characteristic in it travels at its own speed and the faster ones run ahead. This module does not derive it. What matters here is why one is needed — the discontinuous alternative is forbidden — and that is a statement Proposition 3 makes without any further machinery.
The ceiling (3.3) is a consequence of (2.4), and (2.4) assumed that no energy leaves the layer. Molecular gas radiates efficiently, and a dense shocked molecular cloud can lose its thermal energy far faster than the gas crosses the layer. Then the correct closure is not "no losses" but "the temperature does not change", and the ceiling disappears.
The numbers are startling and they matter for the rest of the book. Molecular hydrogen at $10\ \mathrm{K}$ with $\mu = 2.33$ has $c_T = 0.189\ \mathrm{km\,s^{-1}}$. A $10\ \mathrm{km\,s^{-1}}$ flow — a very ordinary speed in the interstellar medium, far below anything a supernova produces — is then at $\mathcal{M} = 52.9$ and compresses the gas by a factor $2802$. Put the same speed through an adiabatic shock and it is at $M_1 = \mathcal{M}/\sqrt\gamma = 41.0$, because the adiabatic sound speed in the same gas is $\sqrt\gamma\,c_T = 0.244\ \mathrm{km\,s^{-1}}$, and it compresses by $3.993$. That is three orders of magnitude, and it is why the dense filaments in which Module 5's Jeans criterion is actually met are made by shocks and not by gravity alone. The whole difference is one closure: whether the layer keeps its heat or radiates it away. ($\gamma = 5/3$ and not $7/5$ is right for H₂ at $10\ \mathrm{K}$: the rotational levels of the molecule have a characteristic temperature near $85\ \mathrm{K}$ and are frozen out at $10\ \mathrm{K}$, leaving only the three translational degrees of freedom.)
$\mathcal{M}$, the isothermal Mach number of (3.5), is written in a different typeface from $M_1$ deliberately. They are ratios to different speeds — $c_T$ is the isothermal sound speed, smaller than the adiabatic $c$ of Definition 2 by $\sqrt\gamma$ — so quoting a Mach number for a radiating shock without saying which one is quoted is an error of a factor $1.29$ at $\gamma = 5/3$. This book uses $M$ for adiabatic and $\mathcal{M}$ for isothermal throughout this module and nowhere else.Deposit energy $E$ at a point at $t = 0$, in a uniform gas of density $\rho_0$ and negligible pressure, and follow the shock that runs outward. The problem is the cleanest use in this book of an argument that needs no equations at all.
Three consequences follow immediately and none needs the interior solution. The shock speed is $v_s = \mathrm{d}R/\mathrm{d}t = \tfrac25 R/t \propto t^{-3/5}$, so the blast decelerates; by Proposition 7 below, which uses nothing from this section, the post-shock temperature goes as $v_s^2 \propto t^{-6/5}$; and, because $R^5t^{-2} = \xi_0^5E/\rho_0$ is a constant of the motion, a single measurement of $R$ at a single $t$ determines $E$ once $\xi_0$ and $\rho_0$ are known. Taylor used exactly that, and Problem K3 shows what it hides.
What dimensional analysis cannot supply is $\xi_0$. That number contains all the dynamics, and getting it requires solving the interior flow.
For (4.3): in the strong-shock limit Proposition 2 gives $\rho_2 = \rho_0(\gamma+1)/(\gamma-1)$, $v_2 = D(\gamma-1)/(\gamma+1)$ and $p_2 = 2\rho_0D^2/(\gamma+1)$, with $D = \mathrm{d}R/\mathrm{d}t = \tfrac25R/t$ supplied by (4.1) itself. Change frame before using $v_2$. Definition 1 measures $v_2$ in the rest frame of the surface, while $v(r,t)$ in (4.2) is a laboratory-frame velocity, in the frame where the unshocked gas ahead is at rest. The two differ by $D$: the shocked gas moves outward at $v_{2,\rm lab} = D - v_2 = D[1 - (\gamma-1)/(\gamma+1)] = 2D/(\gamma+1)$, which is $D/2$ at $\gamma = 5/3$ while $v_2$ itself is $D/4$. Evaluating (4.2) at $r = R$ and dividing gives $u(1) = v_{2,\rm lab}\,t/R = \tfrac25\cdot 2/(\gamma+1)$, $g(1) = \rho_2/\rho_0$ and $\Pi(1) = p_2/[\rho_0(R/t)^2] = 2(4/25)/(\gamma+1)$, which are (4.3). So §4 introduces no new physical input: the boundary condition of the blast wave is Proposition 2 plus similarity.
For (4.4): the total energy inside the shock is $$ E = \int_0^R\left[\tfrac12\rho v^2 + \frac{p}{\gamma-1}\right]4\pi r^2\,\mathrm{d}r = 4\pi\rho_0\frac{R^5}{t^2}\,J $$ after substituting (4.2) and $r = \lambda R$. But $R^5/t^2 = \xi_0^5E/\rho_0$ by (4.1), so $E$ cancels from both sides and what is left is $1 = 4\pi\xi_0^5J$. The energy integral does not determine the energy; it determines the constant. ∎
Two independent verifications are worth stating, because a constant obtained by numerical integration of an eliminated system is exactly the sort of number that can be wrong without looking wrong.
First, internal. The elimination that produced $u'$ used two of the three equations and discarded them. Differentiating the integrated solution numerically and substituting it back into the discarded continuity and entropy equations gives residuals no larger than $7\times10^{-9}$ anywhere between $\lambda = 0.1$ and $\lambda = 0.9$, for both values of $\gamma$. That tests the algebra, not merely the convergence of the integrator.
Second, external. Tang & Chevalier (2017) state that for $\gamma = 5/3$ the constant in $R = (\xi Et^2/\rho)^{1/5}$ equals $2.026$, citing Taylor (1946), Sedov (1959) and Book (1994); that is $\xi_0 = 2.026^{1/5} = 1.151670$ against $1.151666$ computed here, a ratio of $0.999997$. Kamm & Timmes (2007) define their spherical standard problem by $\rho_0 = 1\ \mathrm{g\,cm^{-3}}$, $\gamma = 1.4$ and $E = 0.851072\ \mathrm{erg}$, which puts the shock at $r = 1.0\ \mathrm{cm}$ at $t = 1.0\ \mathrm{s}$; that statement is the constant, $\xi_0 = 0.851072^{-1/5} = 1.032777$, and it agrees to all six figures.
The interior of the solution is as instructive as the constant. Nearly all the shocked mass sits in a thin shell just behind the front: for $\gamma = 5/3$ the mass median radius is $r/R = 0.9392$, the density at $r/R = 0.5$ has fallen to $1.53$ per cent of its post-shock value, and the pressure has flattened to a plateau at $0.306$ of the post-shock pressure and stays there all the way in. A hot, nearly empty, nearly uniform-pressure interior driving a thin dense shell — that is the picture, and §7 shows it becoming literally true as the snowplough phase arrives.
Every module in this book is checked against a published measurement, and the checks have been of uneven quality. Module 3's solar comparison was partly circular, because the model it tested had gone into the model that produced the "measurement". Module 7's laboratory check carried a standard deviation but its astronomical half carried none. This module's anchor has neither defect, and it is worth saying why before any number appears.
Taylor's Table 1 is twenty-five radii read off photographic film with a length scale in the frame, against twenty-five shutter times read off the same film. Nothing about the blast-wave solution went into producing those numbers. No one solved (4.1) to obtain them and no fitted model stands between the film and the table. So the table can test the theory, which is precisely what Module 3's solar model could not do. The prediction, moreover, was published in 1941 — four years before the explosion, nine years before Taylor saw the film — so there was no opportunity to tune it.
The weaknesses are worth stating with equal directness. The radii are read by eye from 1945 photographs and carry no quoted uncertainty. The times come from the camera. Taylor labels every row with the source of its photograph, his "authority" column, in four groups — A, rows 1–14, a strip of small images from MDDC-221; B, rows 15–20, declassified photographs lent by the Ministry of Supply; C, rows 21–23, a second strip of small images from MDDC-221; D, rows 24–25, large single photographs from MDDC-221 — and the source turns out to matter. And the reader is asked to believe a table transcribed from a 1950 letterpress page; §5.1 closes that gap with a check of the data file against the paper's own logarithm columns.
Fit $\log_{10}R$ against $\log_{10}t$ by least squares and compare the slope with the $2/5$ that Proposition 5 forces. Deciding which rows belong in the fit is a judgement, so all three defensible choices are reported.
| rows used | $t$ range | fitted exponent | ratio to $2/5$ | departure |
|---|---|---|---|---|
| all 25 | 0.10–62.00 ms | $0.4058 \pm 0.0076$ | $1.0146$ | $+0.8\,\sigma$ |
| 24, from 0.24 ms (Taylor's own range) | 0.24–62.00 ms | $0.3904 \pm 0.0024$ | $0.9759$ | $-3.9\,\sigma$ |
| 19, rows 2–20 | 0.24–4.61 ms | $0.3989 \pm 0.0049$ | $0.9972$ | $-0.2\,\sigma$ |
Read the last column as a warning, not as a result. The three fitting choices do not agree with one another inside their formal errors: $+0.8\sigma$, $-3.9\sigma$ and $-0.2\sigma$ cannot all be describing the same quantity with correctly estimated uncertainties. Those errors assume the scatter about the line is independent noise. §5.2 shows that it is not, and that is a finding about the model rather than a defect of the fit.
The ratio $0.9759$ is $2.408$ per cent from unity, and is quoted above as "within 2.4 per cent". That is the largest departure of the three, and it must not be restated to a third digit anywhere, because the fit's own third digit is not stable against the choice of rows.The data file is checked against the paper, not against a transcription of it. Taylor prints $\log_{10}t$, $\log_{10}R$ and $\tfrac52\log_{10}R$ beside his $t$ and $R$ columns, and all twenty-five $R$ values in afd/data/taylor1950_trinity.dat reproduce his $\log_{10}R$ column to its last printed digit with $R$ in centimetres. Two times were ambiguous in the scan and were recovered from his own logarithm column: row 8, where the digits read "1·08" and his $\log_{10}t = \overline{3}.033$ confirms $1.079$ ms; and row 12, where the digits read "1·05", which is wrong and also not monotonic with its neighbours, and his $\log_{10}t = \overline{3}.217$ gives $1.648$ ms. A further check is arithmetic: the mean of $\tfrac52\log_{10}R - \log_{10}t$ over his 24 rows is $11.9159$ against the $11.915$ of his own equation (1), which reproduces his printed intercept to four decimal places. That is a check of the data file, not of the physics.
Plot the residual of each row from the fixed-slope line — that is, the quantity $\tfrac52\log_{10}R - \log_{10}t$, which Proposition 5 says should be the same number for every row — and the scatter is not random. Group the rows in four blocks in time. The three later blocks are chosen to follow Taylor's authority column, with one departure that must be stated because it matters below: the first row is separated from the rest of authority A, because it is the row the model is expected to fail at, and authorities C and D are pooled into one block of five, because D has only two rows:
| rows | $t$ range | mean residual | scatter within the block |
|---|---|---|---|
| row 1 alone | 0.10 ms | $-0.3026$ | — |
| 2–14 | 0.24–1.93 ms | $+0.0150$ | $0.0196$ |
| 15–20 | 3.26–4.61 ms | $-0.0043$ | $0.0048$ |
| 21–25 | 15–62 ms | $-0.0339$ | $0.0074$ |
The three later block means fall monotonically. The step from the second block to the fourth is $0.0490$ in $\tfrac52\log_{10}R$, which is a factor $1.046$ in $R$ at fixed $t$ — 4.6 per cent. The standard errors of those two block means are $0.0056$ and $0.0037$, so the step is 7.3 times their combined error. A structured residual is a failed model, not a noisy measurement.
One honest confound, and it is stated before the conclusion because it limits the conclusion. The block boundaries coincide with changes of photographic source, so the 7.3σ step between block 2 and block 4 could in principle be an instrumental offset rather than a change of physics, and this table cannot separate the two. Taylor quotes no uncertainty on any radius, so there is no measurement error to compare an instrumental offset against. The physical mechanism offered below — the neglected ambient pressure doing work on the flow after about 5 ms — predicts the observed sign and this module computes no size for it, so it cannot discriminate either. What does not depend on the source boundaries is the structure inside authority A: rows 2–14 scatter about $+0.0150$ while row 1, from the same strip of film, sits $0.32$ below them, sixteen times their internal scatter. Rows 21–23, all authority C, also fall steadily, $-0.0237 \to -0.0290 \to -0.0440$, though three points in order is weak evidence taken alone. The refutation below rests on those two, on the whole monotone ordering, and on the Mach-number decline computed in the next list — not on the block step by itself.
What fails, and when, can be read off the same table, and each item below is a hypothesis of Definition 1 or Proposition 5 breaking in a specific way.
All of this is one figure. Fig. 3 shows the twenty-five points on the slope-1 line in its upper right panel, where the confirmation of §5.1 is visible and the structure of this section is not — every point sits within a point-width of the line. The lower panel of the same figure plots the residual, by photographic source, with the block means drawn, and the refutation is there and nowhere else. The confirmation and the refutation are the same twenty-five numbers, plotted twice. That is the lesson Module 2 learned about residual plots, applied here to a case where it decides the verdict.
Because $R^5t^{-2}$ is a constant of the motion, the fit determines it, and $E = K\rho_0R^5t^{-2}$ with $K = 1/\xi_0^5$ from Proposition 6 turns it into an energy. The fixed-slope fit over Taylor's own 24 rows gives $R^5t^{-2} = 6.7902\times10^{23}\ \mathrm{cm^5\,s^{-2}}$, and with the exact $K = 0.851072$ for $\gamma = 1.40$ and his $\rho_0$,
$$ E \;=\; 7.2237\times10^{20}\ \mathrm{erg} \;=\; 17.27\ \mathrm{kt}, \qquad 1\ \mathrm{kt} \equiv 4.184\times10^{19}\ \mathrm{erg}. $$Four numbers must be laid beside it, and the arithmetic that relates them is not optional.
| value | source | in modern kt |
|---|---|---|
| $7.14\times10^{20}$ erg, "16,800 tons" | Taylor (1950), his Table 3, $\gamma = 1.40$ | $17.07$ |
| $7.2237\times10^{20}$ erg | this book, same data, exact $K$ | $17.27$ |
| 21 kt | Department of Energy released value | $21.0$ |
| $24.8 \pm 2$ kt | Selby et al. (2021), radiochemistry of trinitite | $24.8 \pm 2.0$ |
Taylor's "16,800 tons" is not 16.8 kt, and the difference is not a rounding. He took one gram of T.N.T. as liberating 1000 calories, $4.18\times10^{10}$ erg, on the long ton, giving $1\ \mathrm{ton} = 4.25\times10^{16}$ erg. The modern kiloton is $4.184\times10^{19}$ erg, so his ton is $1.0158$ modern tons and his 16,800 tons is $17.07$ modern kt. Quoting "16.8 kt" from his paper, as is often done, is an error of 1.6 per cent introduced by the reader rather than by Taylor.
An inconsistency inside the source, reproduced here because it propagates. Taylor's equation (1) gives the intercept $\tfrac52\log_{10}R - \log_{10}t = 11.915$, which implies $R^5t^{-2} = 10^{2\times11.915} = 6.7608\times10^{23}$. His equation (3), two paragraphs later, prints $6.67\times10^{23}$. The two differ by 1.36 per cent, and it is his equation (3) that feeds his energy. Both numbers were read from the page.The direction of the remaining error is known, and it widens the gap. The Trinity site is a desert well above sea level, so its air was less dense than the $1.25\times10^{-3}\ \mathrm{g\,cm^{-3}}$ Taylor assumed. $E$ is linear in $\rho_0$, so a more honest $\rho_0$ makes the recovered energy smaller and the shortfall larger. No elevation is quoted here, because none was read from a source. The statement is about a sign, and a sign needs no number.
The $\pm 2$ kt attached to Selby's 24.8 kt is theirs, from a Bayesian treatment folding thirteen trinitite samples with radiochemical and physics simulation tools; it is not a pure measurement uncertainty. The DOE's 21 kt carries no published uncertainty at all. Selby write that their result "overlaps with the previous DOE value of $21 \pm 2$ kt, where we assigned a conservative 10% relative uncertainty on its value", and that the source of the DOE figure "is not cited in the official DOE publications". The $3.8\,\sigma$ above therefore uses Selby's $\sigma$ and not the DOE's, because the DOE has none.Every shock in this book so far has been inferred from a distance. In 2007 a spacecraft flew through one and measured it from the inside. Voyager 2's plasma experiment recorded a decrease in solar-wind speed from about 9 June 2007, culminating in several crossings of the solar-wind termination shock at a heliocentric radius of 84 AU between 30 August and 1 September 2007.
The paper reporting it supplies something rarer than the crossing itself: a control. The same instrument on the same spacecraft crossed Neptune's inbound bow shock in August 1989, eighteen years earlier, at 30 AU. So the comparison below is not merely theory against observation; it is one shock against another, through one instrument, with the theory saying what should differ and what should not.
From the figure caption comparing the two crossings, read directly: at Neptune's bow shock the solar-wind speed fell by a factor of four and the density increased by a factor of four, while the temperature increased by a factor of 100.
Fig. 4 draws all five comparisons of this section on one logarithmic axis, as measured divided by predicted, so that a confirmation and a refutation are the same kind of mark in the same place.
The factor of four in density is the ceiling (3.3) for $\gamma = 5/3$, and $\gamma = 5/3$ is not fitted — it is fixed by the gas being monatomic. Ratio of measured to predicted: $1.0$. That the density jump and the speed drop are the same number is mass conservation, (2.2), holding to the precision of the read — provided the bow shock is standing in the frame the speeds were measured in, which a planetary bow shock does to that precision, because (2.2) is a statement in the rest frame of the surface and the speeds are measured in the spacecraft's.
The temperature supplies a second, independent test of the same shock. A temperature ratio of 100 implies, by inverting (3.2) at $\gamma = 5/3$, an upstream Mach number $M_1 = 17.8$. That Mach number then predicts a density ratio of $3.96$ through the first of (3.1), against the measured 4. Consistency ratio $0.99$. One Mach number, two independent Rankine–Hugoniot relations, one answer: that is the shape of a real test, and it is what Proposition 2 is for.
At the termination shock the same instrument found the temperature increasing by a factor of only ten and the density by only two. The authors state the consequence plainly: if all the flow energy were transferred to the thermal ions, as happens at planetary bow shocks, the heliosheath temperature would be $10^6$ K, whereas the observed temperature is $10^5$ K.
That $10^6$ K is not taken on trust; Proposition 7 reproduces it. The same paper reports the solar wind decelerating "from 400 to 300 km s$^{-1}$ in three steps" ahead of the crossing. Putting $v_s = 300\ \mathrm{km\,s^{-1}}$ and $\mu = 0.5$, for a fully ionised hydrogen plasma, into (7.1) at $\gamma = 5/3$ gives $T_2 = 1.01\times10^{6}$ K, which is their figure to within 2 per cent. So Check 5 below is this book's prediction against their measurement and not their prediction against their measurement.
One frame caveat, stated because it is of the size of the effect. The $300\ \mathrm{km\,s^{-1}}$ is measured in the spacecraft frame, while (7.1) needs the upstream speed in the frame of the shock, and their Table 1 gives the shock moving outward at $94.0 \pm 3.4\ \mathrm{km\,s^{-1}}$. Using $300 - 94 = 206\ \mathrm{km\,s^{-1}}$ instead gives $T_2 = 4.8\times10^{5}$ K. The agreement with their $10^6$ K is therefore an agreement with their choice of frame, and this module does not claim the shock-frame value is what they used. The refutation survives either choice: the observed $10^5$ K is short by a factor 10 against the first and a factor 4.8 against the second.The authors' explanation is what makes this a refutation of a specific hypothesis rather than a puzzle. The energy per proton falls by 80 per cent across the shock, and it goes to pickup ions — hot protons created when neutral interstellar atoms are ionised inside the heliosphere — rather than to the thermal plasma. A single-fluid ideal gas has nowhere to put that energy, so it predicts a hot thermal population that is not there. And Module 1 has already said why this is possible: the solar wind at 84 AU is collisionless, so no mechanism forces one temperature on all the ion populations. The failure was licensed by Module 1 and is observed here.
The same paper's Table 1 gives, for the second crossing, a compression ratio of $2.38 \pm 0.14$ against an upstream fast-mode Mach number of $4.9 \pm 0.1$. The gas-dynamic prediction of (3.1) at $M_1 = 4.9$ and $\gamma = 5/3$ is $3.556$, so the measurement is $0.669$ of it — a departure of 8.4 formal standard deviations, which would be by far the largest refutation in this book. It is not used, for two reasons.
First, it is partly circular. The table's caption states that the shock orientation is found as "the direction that minimizes the differences of the Rankine–Hugoniot conditions across the shock", and that the shock speed, the field angle and the Mach numbers are then calculated from that orientation. The compression ratio and the Mach number therefore come from one Rankine–Hugoniot fit to one data window, so comparing them against the Rankine–Hugoniot prediction tests the fit against itself. This is Module 3's circularity in a new costume, and it is refused for the same reason.
Second, it is the wrong formula. The same table gives the angle between the shock normal and the magnetic field as $82.8^\circ \pm 3.9^\circ$: this is a quasi-perpendicular magnetohydrodynamic shock, and (3.1) is a gas-dynamic result. The right formula is the oblique magnetohydrodynamic jump condition, which this book does not derive anywhere. Module 12 builds magnetohydrodynamics — flux freezing, magnetic pressure and tension, Alfvén waves, the magnetorotational instability — and stops short of shock jump conditions. Saying so is better than promising a repayment that no planned module makes.
The temperature comparison of §6.2 has neither defect. The observed $10^5$ K and the predicted $10^6$ K are separate statements: the first is a measurement, the second is what full thermalisation of the measured flow energy would give, and no fit connects them.
The Sedov solution is exact for a point explosion in a pressureless uniform gas, and a supernova remnant is neither at the start nor at the end of its life. This section puts brackets around the interval in which (4.1) is the right description, and the brackets are as much a result as the solution is.
Take the standard numbers: $E = 10^{51}$ erg, ambient hydrogen density $n_{\rm H} = 1\ \mathrm{cm^{-3}}$, so $\rho_0 = 1.4\,m_u n_{\rm H} = 2.325\times10^{-24}\ \mathrm{g\,cm^{-3}}$ with helium at cosmic abundance, $\mu = 0.61$ for the fully ionised gas behind the shock, and $\gamma = 5/3$.
With $\xi_0 = 1.15167$ from Proposition 6, (4.1) and (7.1) give the remnant's whole adiabatic history:
| age (yr) | $R$ (pc) | $v_s$ (km s⁻¹) | $T_2$ (K) | swept mass ($M_\odot$) |
|---|---|---|---|---|
| 100 | 1.988 | 7774 | $8.31\times10^{8}$ | 1.1 |
| 300 | 3.085 | 4021 | $2.22\times10^{8}$ | 4.2 |
| 1 000 | 4.993 | 1953 | $5.25\times10^{7}$ | 17.9 |
| 3 000 | 7.748 | 1010 | $1.40\times10^{7}$ | 66.9 |
| 10 000 | 12.541 | 490.5 | $3.31\times10^{6}$ | 283.8 |
| 30 000 | 19.462 | 253.7 | $8.86\times10^{5}$ | 1 061 |
| 100 000 | 31.502 | 123.2 | $2.09\times10^{5}$ | 4 498 |
$R \propto t^{2/5}$, $v_s \propto t^{-3/5}$ and $T_2 \propto v_s^2 \propto t^{-6/5}$: the third follows from the second by (7.1) and needs no new input.
The cooling function is stated, not fitted, in two branches per unit $n_en_{\rm H}$: $$ \Lambda(T) \;=\; \begin{cases} 2.1\times10^{-27}\sqrt{T} & T \gt 10^7\ \mathrm{K} \quad\text{(thermal bremsstrahlung)}\\[2pt] \Lambda_7\,(T/10^7)^{-0.7} & 10^4 \lt T \le 10^7\ \mathrm{K} \quad\text{(collisionally excited lines)}\end{cases} \tag{7.2}$$ with $\Lambda_7 = 2.1\times10^{-27}\sqrt{10^7}$ so that the branches join continuously, and $$ t_{\rm cool} \;=\; \frac{\tfrac32 n_{\rm tot}k_BT}{n_en_{\rm H}\Lambda(T)}, \qquad n_{\rm tot} = \frac{\rho}{\mu m_u}, \quad n_e = 1.2\,n_{\rm H}. \tag{7.3}$$ ∎
A one-line criterion is not a simulation, and the honest thing is to say by how much it misses. Cioffi, McKee & Bertschinger (1988) computed the radiative transition with a hydrodynamic code and fitted the result. Their equations (3.33a) and (3.33b) give, for solar abundances and $E_{51} = n_0 = 1$,
$$ R_{\rm PDS} = 14.0\ \mathrm{pc}, \qquad v_{\rm PDS} = 413\ \mathrm{km\,s^{-1}}, \qquad t_{\rm PDS} = \tfrac25 R_{\rm PDS}/v_{\rm PDS} = 1.326\times10^{4}\ \mathrm{yr}. $$ $t_{\rm PDS}$ above is derived from their two fitted quantities using $R = \tfrac25v_st$, which holds at the end of the Sedov stage. It can be checked against the same paper by a second route: their equation (3.10) gives a shell-formation time $t_{\rm sf} = 3.61\times10^4$ yr for these parameters, and their equation (3.11) sets $t_{\rm PDS} = t_{\rm sf}/e$, which is $1.328\times10^4$ yr. The two routes agree to 0.15 per cent.Comparing:
After $t_{\rm rad}$ the shell is thin, dense and driven by the pressure of the hot interior — the interior that Proposition 6 showed is nearly empty and nearly uniform in pressure. Then $R \propto t^{0.30}$ rather than $t^{0.40}$, and the similarity solution of §4 no longer applies. That phase is not derived in this book.
Module 7 closed by recording a gap: its Rayleigh–Taylor half was checked against a laser experiment and against nothing in the sky, and no module from 1 to 10 was going to repay it. This section repays it, using the structure that §7 has just built, and the answer is a refutation.
A young supernova remnant is not one shock but three surfaces. The ejecta drive a forward shock (the blast wave) into the interstellar medium. The ambient gas pushes back on the ejecta through a reverse shock running into them. Between the two lies the contact discontinuity, separating shocked ejecta from shocked ambient gas. Shocked ejecta are the heavier fluid and sit inside; shocked ambient gas is lighter and sits outside; and the shell decelerates. In the frame of the decelerating interface that is exactly Module 7's unstable ordering, and the contact discontinuity grows Rayleigh–Taylor fingers.
Module 7 measured a mixing law, $h = \alpha A g t^2$, for a plane interface under a constant acceleration, and derived $g = m(1-m)R/t^2$ for a shell whose radius follows $R \propto t^m$. Putting the second into the first is the whole of this section's theory.
What Proposition 9 produces is a bound rather than a value, and that is what makes §8 a test and not a fit: if the observed width exceeds the bound, no choice of the two free quantities can rescue the prediction.
Two consequences are worth stating before any number is put in. No distance and no age are needed: (8.1) is a ratio of radii, so the disagreement among published distances to Tycho — which is real — cannot affect it, and neither can the epoch at which the remnant is observed. And (8.2) turns the comparison into a test rather than a fit: if the observed width exceeds $\alpha/4$ times the radius ratio, no choice of $A$ or $m$ can rescue the prediction.
The inputs, each read from the paper that supplies it:
Putting those into (8.1):
$$ \frac{h}{R_{\rm BW}} \;=\; 0.038 \times 1 \times 0.2496 \times 0.77 \;=\; 0.0073, \qquad\text{0.73 per cent of the blast-wave radius,} $$with the $\pm0.008$ on $\alpha_B$ alone spanning $0.58$ to $0.88$ per cent, and the ceiling (8.2) at $0.73$ per cent for $\alpha_B$ and $0.89$ per cent one standard deviation above it.
Against that, four rungs, every one from a paper read in full:
| $R/R_{\rm BW}$ | what it is | source |
|---|---|---|
| 0.77 | one-dimensional contact discontinuity, pure hydrodynamics | Warren et al. (2005), their Sect. 7, who state it; the number is not printed in Wang & Chevalier's text, so it is their reading of that paper's results |
| 0.85 | two-dimensional Rayleigh–Taylor finger tips, pure hydrodynamics | Wang & Chevalier (2001), their own text at $t' = 1.6$; the same paper puts Tycho at $t' = 1.7$, and the finger extent is not restated there |
| 0.93 | Tycho's mean contact discontinuity, deprojected | Warren et al. (2005), abstract |
| 1.00 | Tycho's ejecta clumps, at several azimuths | Warren et al. (2005), their Sect. 3 |
Compare a width with a width, never with a position. Every observed width below is measured outward from the same place, the one-dimensional contact discontinuity at $0.77$:
| observed width | value | predicted ÷ observed |
|---|---|---|
| to the simulated finger tips | $0.08\,R_{\rm BW}$ | $0.091$ |
| to Tycho's mean contact discontinuity | $0.16\,R_{\rm BW}$ | $0.046$ |
| to Tycho's farthest clumps | $0.23\,R_{\rm BW}$ | $0.032$ |
And nothing inside the law can close the gap. To reach even the simulated $0.08\,R_{\rm BW}$, (8.1) would need $\alpha = 0.416$: eleven times the measured $0.038 \pm 0.008$, and 8.3 times the largest theoretical value Module 7 found anywhere, $\sim 0.05$ for three-dimensional immiscible fluids. $A$ cannot help, because it is already at its ceiling of 1. $m$ cannot help, because $m(1-m)$ is already within 1 per cent of its ceiling of $\tfrac14$.
Fig. 6 draws the predicted width and the three observed ones on one radial axis, at one scale, every bar starting from the same radius. The prediction is the hairline.
Refuted: the transfer of a plane-interface, constant-acceleration, two-uniform-fluid, forgotten-seed mixing law to a decelerating supernova shell. Not refuted: the law itself, which Module 7 checked in the regime in which it was measured and which is not under test here.
Three reasons the transfer fails, in order of size, and the module claims no more precision for them than it has:
What this section does not use, and why. Warren's strongest single azimuthal mode of the contact discontinuity, $n = 6$, has an amplitude of 3.3 per cent of the mean radius — a tempting comparator, and the right order of magnitude to make the prediction look better. It is refused, because the same sentence records that "the BW shows nearly the same value", and the blast wave is not Rayleigh–Taylor unstable. A number that both surfaces share cannot be measuring an instability that only one of them has. What is specific to the contact discontinuity is the high-wavenumber power, where it carries nearly an order of magnitude more than the blast wave, and its spectral slope, $P \propto k^{-1.5}$ against $k^{-2.2}$. Those are qualitative, and they are quoted as qualitative.
Every idealisation used above is named here, with the module that repays it where one exists and a plain statement of "not repaid in this book" where none does.
One thing this module does not miss is worth recording, because it is unusual. The check in §5 is against a table that no model touched, and the check in §6 is against an instrument that flew through the object. Neither has the circularity that Module 3 had to declare. What they have instead is a shortage of quoted uncertainties, and that is why §5.3 leans on Selby's $\sigma$ — the only honest standard deviation in the whole Trinity comparison.
Nine problems in the house set: three conceptual (C), three derivational (D), three computational (K). Every numerical answer below is produced by afd/figs/m08_numbers.py and independently verified by afd/figs/m08_problems_check.py.
(7.1) at $\gamma = 5/3$ is $T_2 = 3\mu m_u v_s^2/(16k_B)$, so $T_2 \propto \mu$ exactly. With $v_s = 5\times10^8\ \mathrm{cm\,s^{-1}}$: $\mu = 0.61$ gives $3.439\times10^8$ K, $\mu = 1.00$ gives $5.638\times10^8$ K, $\mu = 0.50$ gives $2.819\times10^8$ K.
The three differ by a factor 2.0 across the range of $\mu$ that a real remnant might have — ionised cosmic abundance, neutral atomic hydrogen, fully ionised pure hydrogen. A post-shock temperature quoted without its $\mu$ is therefore uncertain by a factor 2 before any measurement error, which is larger than most of the observational uncertainties it would be compared against.
At $M_1 = 8$, $\gamma = 5/3$: $\rho_2/\rho_1 = 3.8209$, which is 95.5 per cent of the ceiling 4; $p_2/p_1 = 79.75$; $T_2/T_1 = 20.87$; $M_2 = 0.4583$. The downstream density is $n_2 = 3.8209\times5 = 19.10\ \mathrm{cm^{-3}}$.
By the exact ratio, $T_2 = 20.87\times10^5 = 2.087\times10^6$ K. By (7.1) with $v_s = 4\times10^7\ \mathrm{cm\,s^{-1}}$ and $\mu = 0.61$, $T_2 = 2.201\times10^6$ K. The two differ by 5 per cent because $M_1 = 8$ is not infinite: (7.1) drops the $-(\gamma-1)$ in the numerator of $p_2/p_1$ and the $+2$ in the denominator of $\rho_2/\rho_1$, and at $M_1 = 8$ those terms are not yet negligible. The strong-shock form is an asymptote, and D1 asks where it becomes safe.
Note also $\Delta s/k_B = 3.2171$ from (3.4) — the entropy jump is of order the Boltzmann constant per particle, which is what "irreversible" means quantitatively here.
At $M_1 = 3$: $\gamma = 5/3$ gives $\rho_2/\rho_1 = 3.0000$, $p_2/p_1 = 11.0000$ and $\Delta s/k_B = 0.8503$; $\gamma = 7/5$ gives $3.8571$, $10.3333$ and $1.1137$.
Below $M_1 = 1$ the conservation laws still return solutions: $M_1 = 0.9$ gives $\rho_2/\rho_1 = 0.8504$, $p_2/p_1 = 0.7625$, $\Delta s/k_B = -0.00159$; $M_1 = 0.7$ gives $0.5616$, $0.3625$, $-0.07970$; $M_1 = 0.5$ gives $0.3077$, $0.0625$, $-1.21225$.
Reason one, thermodynamic. Every one of those entropy jumps is negative, so every one is forbidden by the second law. This applies on the whole range $M_1 \lt 1$.
Reason two, and it is not thermodynamic. Below $M_1 = \sqrt{(\gamma-1)/2\gamma} = 0.4472$ for $\gamma = 5/3$, the formula for $p_2/p_1$ returns a negative pressure — $-0.0500$ at $M_1 = 0.4$. On that range the rarefaction shock is not a solution the second law rejects; it is not a solution at all. The two reasons cover different ranges and must not be merged.
Solve $\rho_2/\rho_1 = 0.99(\gamma+1)/(\gamma-1)$ for $M_1$ using the first of (3.1). For $\gamma = 5/3$: $M_1 = 17.2$, where the ratio is $3.9600$ against the ceiling $4.0000$. For $\gamma = 7/5$: $M_1 = 22.2$, where it is $5.9400$ against $6.0000$. The diatomic gas needs a stronger shock because its ceiling is further away in ratio terms.
Trinity at 1 ms: from Taylor's table $R \approx 38\ \mathrm{m}$, so $v_s = \tfrac25R/t = 14.4\ \mathrm{km\,s^{-1}}$. The ambient sound speed in air at one atmosphere with Taylor's $\rho_0$ is $0.337\ \mathrm{km\,s^{-1}}$, giving $M_1 = 43$. Strong by any measure, and comfortably past the 22.2 that $\gamma = 7/5$ requires.
Air. Adding viscosity gives the layer a width of a few mean free paths. In sea-level air $\lambda_{\rm mfp} \sim 6.5\times10^{-6}$ cm, so the layer is $\sim 2\times10^{-5}$ cm. The Trinity fireball at 1 ms had a radius of 38 m $= 3.8\times10^3$ cm, so the layer is $2\times10^{8}$ times thinner than the object it bounds. Definition 1 is safe by eight orders of magnitude.
The termination shock. The measured layer widths were 100 000–300 000 km, described by the observers as "a few times the ion inertial length", so $10^{10}$–$3\times10^{10}$ cm. The collisional mean free path for proton–proton Coulomb collisions, $\lambda_{\rm mfp} = (k_BT)^2/(\pi n e^4\ln\Lambda)$ with $n \approx 2\times10^{-3}\ \mathrm{cm^{-3}}$ and $T \approx 10^5$ K read approximately off the published figure axes and $\ln\Lambda = 25$, is $2.28\times10^{16}\ \mathrm{cm} = 1524$ AU.
The layer is therefore $8\times10^{5}$ to $2\times10^{6}$ times thinner than a mean free path. That is what "collisionless shock" means, and it is why the one-fluid closure fails there while the conservation laws do not: Proposition 1 needs only that fluxes balance, which holds whatever mediates the transition, but the assumption that one temperature describes all the ions needs collisions, and there are none.
Integrate $\lambda^2 g(\lambda)$ outward and find where it reaches half its total. For $\gamma = 5/3$ the mass median is at $r/R = 0.9392$; for $\gamma = 7/5$ at $r/R = 0.9581$. Nearly all the mass is in a thin shell just behind the front.
The interior is correspondingly empty. For $\gamma = 5/3$, $\rho/\rho_2$ is $0.3563$ at $r/R = 0.9$, $0.01531$ at $0.5$ and $3.4\times10^{-10}$ at $0.01$; for $\gamma = 7/5$, $0.2054$, $0.00172$ and $3.1\times10^{-16}$.
The pressure does the opposite: $p/p_2$ is $0.3098$ at $r/R = 0.5$ and $0.3062$ at $r/R = 0.01$ for $\gamma = 5/3$ — flat. That is what it must be. A region with almost no mass in it cannot support a pressure gradient, because there is nothing for the gradient to accelerate. The snowplough phase of §7 is the limit of this picture: a thin dense shell driven by a uniform interior pressure.
Invert (4.1) for $t$ at fixed $R$: $t = (R/\xi_0)^{5/2}\sqrt{\rho_0/E}$.
| $n_{\rm H}$ (cm⁻³) | age (yr) | $v_s$ (km s⁻¹) | $T_2$ (K) |
|---|---|---|---|
| 0.1 | $1.016\times10^{4}$ | 770 | $8.16\times10^{6}$ |
| 1.0 | $3.212\times10^{4}$ | 244 | $8.16\times10^{5}$ |
| 10 | $1.016\times10^{5}$ | 77 | $8.16\times10^{4}$ |
At fixed radius $t \propto \rho_0^{1/2}$, so a factor 100 in density is only a factor 10 in the inferred age. The square root is doing useful work: an order-of-magnitude guess at the ambient density costs a factor $\sqrt{10} = 3.2$ in the age, which is often tolerable. The temperature is less forgiving, falling by the full factor 100, because $T_2 \propto v_s^2 \propto t^{-6/5}$.
| $E$ (erg) | $n_{\rm H}$ (cm⁻³) | $t_{\rm rad}$ (yr) | $R_{\rm rad}$ (pc) | $T_2$ (K) | $v_s$ (km s⁻¹) |
|---|---|---|---|---|---|
| $10^{51}$ | 1.0 | $4.085\times10^{4}$ | 22.02 | $6.11\times10^{5}$ | 211 |
| $10^{51}$ | 0.1 | $1.458\times10^{5}$ | 58.06 | $3.34\times10^{5}$ | 156 |
| $10^{51}$ | 10 | $1.144\times10^{4}$ | 8.35 | $1.12\times10^{6}$ | 285 |
| $10^{50}$ | 1.0 | $2.441\times10^{4}$ | 11.31 | $4.52\times10^{5}$ | 181 |
Density matters more. A factor 100 in $n_{\rm H}$ moves $t_{\rm rad}$ by a factor 12.7 and $R_{\rm rad}$ by 7.0; a factor 10 in $E$ moves them by only 1.67 and 1.95. The reason is that cooling is a two-body process, so the cooling rate per unit volume goes as $n^2$ while the energy available goes as $E$ once.
Note also that the post-shock temperature at the transition is not a constant of the problem: it ranges from $3.34\times10^5$ to $1.12\times10^6$ K across these four cases. Quoting "the remnant becomes radiative at $10^6$ K" as though it were universal is wrong by a factor 3 at the low-density end.
A single $(t, R)$ pair gives $R^5t^{-2}$ directly, and $E = K\rho_0R^5t^{-2}$ with $K = 0.851072$ and $\rho_0 = 1.25\times10^{-3}\ \mathrm{g\,cm^{-3}}$:
| row | $t$ (ms) | $E$ (kt) |
|---|---|---|
| 1 | 0.10 | 4.28 |
| 4 | 0.52 | 18.63 |
| 11 | 1.50 | 19.50 |
| 15 | 3.26 | 17.10 |
| 20 | 4.61 | 16.52 |
| 25 | 62.00 | 14.33 |
The estimates run from 4.28 to 19.50 kt. Excluding row 1 they still run from 14.33 to 19.50, a spread of 36 per cent, and they fall monotonically with time after row 11.
That spread is the refutation of §5.2, seen from the other side. If the model were right, every row would give the same energy. A one-point estimate hides the disagreement completely and returns a number with no visible uncertainty — which is precisely what makes single-point energy estimates dangerous. The same data, plotted as a residual, shows the structure at $7.3\,\sigma$.
Every symbol used in this module, with its units and the place it is defined. Three symbols collide with earlier modules and the collisions are resolved here rather than left to the reader.
| symbol | meaning | units | defined |
|---|---|---|---|
| $\rho_1, v_1, p_1$ / $\rho_2, v_2, p_2$ | density, normal speed and pressure upstream / downstream, in the frame of the shock | $\mathrm{g\,cm^{-3}}$, $\mathrm{cm\,s^{-1}}$, $\mathrm{dyn\,cm^{-2}}$ | Definition 1 |
| $D$, $v_s$ | speed of the shock surface through the laboratory frame; $v_s$ is the same quantity when the upstream gas is at rest, which is the case throughout §§4–8 | $\mathrm{cm\,s^{-1}}$ | Definition 1, (4.1) |
| $w$ | specific enthalpy $\gamma p/[(\gamma-1)\rho]$. Written $w$ and not $h$, because $h$ is the mixing width of §8 and this module uses both | $\mathrm{erg\,g^{-1}}$ | Definition 1, (2.4) |
| $M_1$, $M_2$ | upstream and downstream Mach numbers, ratios to the adiabatic sound speed | — | Definition 2, (3.2) |
| $\mathcal{M}$ | the isothermal Mach number $v_1/c_T$ of Proposition 4. A different typeface from $M$ because it is a ratio to a different speed, smaller by $\sqrt\gamma$ | — | (3.5) |
| $\gamma$, $c$, $c_T$ | ratio of specific heats; adiabatic sound speed $\sqrt{\gamma p/\rho}$; isothermal sound speed $\sqrt{k_BT/(\mu m_u)}$ | —, $\mathrm{cm\,s^{-1}}$, $\mathrm{cm\,s^{-1}}$ | (2.1), (3.5) |
| $\Delta s$ | entropy change per particle across the shock, quoted as $\Delta s/k_B$ | $\mathrm{erg\,K^{-1}}$ | (3.4) |
| $E$, $\rho_0$ | energy deposited at the point explosion; uniform ambient density ahead of the blast wave | erg, $\mathrm{g\,cm^{-3}}$ | Proposition 5 |
| $\lambda$ | COLLISION, declared. In §4 and Problem D3, $\lambda = r/R$ is the similarity variable, dimensionless. In Module 1 and in Problem D2 only, $\lambda$ with a subscript — $\lambda_{\rm mfp}$ — is a mean free path in cm; in Modules 4–7 a bare $\lambda$ is a wavelength. No section of this module uses two of the three. Problem D2 uses the mean free path and contains no similarity variable; §4 uses the similarity variable and contains no wavelength, and §7 uses neither, its $\Lambda(T)$ being a different symbol | —, cm | (4.2) |
| $u$, $g$, $\Pi$ | the similarity profiles of velocity, density and pressure in (4.2). $\Pi$ rather than $h$, for the reason given against $w$ above; the generating script m08_numbers.py calls it h, and the two are the same function | all dimensionless | (4.2) |
| $\xi_0$, $J$, $K$ | the Sedov constant of (4.1); the energy integral of (4.4); Taylor's constant $K = 1/\xi_0^5 = 4\pi J$ | all dimensionless | (4.1), (4.4) |
| $T_2$, $\mu$ | post-shock temperature; mean molecular weight measured against $m_u$, book-wide from Module 6. $\mu$ must be quoted with every $T_2$ | K, — | (7.1) |
| $\Lambda(T)$, $t_{\rm cool}$, $n_e$, $n_{\rm H}$ | radiative cooling coefficient per unit $n_en_{\rm H}$; cooling time; electron and hydrogen number densities, with $n_e = 1.2\,n_{\rm H}$ for the fully ionised gas | $\mathrm{erg\,cm^3\,s^{-1}}$, s, $\mathrm{cm^{-3}}$ | (7.2), (7.3) |
| $t_{\rm rad}$, $R_{\rm rad}$ / $t_{\rm PDS}$, $R_{\rm PDS}$, $t_{\rm sf}$ | the radiative transition from the criterion of Proposition 8 / the same transition from Cioffi et al.'s fitted formulae, and their shell-formation time. The two pairs are different quantities and §7.1 compares them rather than identifying them | yr, pc | Proposition 8, §7.1 |
| $m$ | COLLISION, declared. In §8 only, $m$ is the expansion index of $R \propto t^m$, dimensionless. Everywhere else in this book $m$ with a subscript is a mass — $m_u$ the atomic mass unit, $m_p$ the proton mass — and $M_\odot$ is the solar mass. §8 contains no mass symbol without a subscript | — | (8.1) |
| $h$, $\alpha_B$, $A$ | the Rayleigh–Taylor mixing width of §8; the bubble mixing coefficient measured in Module 7; the Atwood number of Module 7 Definition 2. $h$ is the mixing width in this module and nothing else — the enthalpy is $w$ and the similarity pressure is $\Pi$ | cm, —, — | (8.1) |
| $R_{\rm BW}$, $R_{\rm cd}$, $R_{\rm RS}$ | radii of the blast wave (forward shock), the contact discontinuity and the reverse shock in a young remnant | cm or arcsec as marked | §8, Fig. 5 |
| $k_B$, $m_u$, $e$, $\ln\Lambda$ | Boltzmann's constant; atomic mass unit; elementary charge in e.s.u.; the Coulomb logarithm of Module 1, used only in Problem D2 | $\mathrm{erg\,K^{-1}}$, g, esu, — | Problem D2 |
Every published value this module compares against, with what was taken from it and from which page. Each was read from the text of the paper, extracted locally from the PDF; none was taken from a summary or from memory. One convention, used throughout this module: a § on its own is a section of this book and is a link; a section of a cited paper is written "their Sect. N", with no § and no link.
afd/data/taylor1950_trinity.dat, whose header records the two times recovered from his logarithm column.